【lightoj1294】数学

墨蓝 2022-07-17 00:27 61阅读 0赞

H - H 使用long long

**Time Limit:**2000MS **Memory Limit:**32768KB **64bit IO Format:**%lld & %llu

Submit [Status][] [Practice][] [LightOJ 1294][] [![12x12_uDebug.png][]uDebug][12x12_uDebug.png_uDebug]

Description

Given two integers: **n** and **m** and **n** is divisible by **2m**, you have to write down the first **n** natural numbers in the following form. At first take first **m**integers and make their sign negative, then take next **m** integers and make their sign positive, the next **m** integers should have negative signs and continue this procedure until all the **n** integers have been assigned a sign. For example, let **n** be **12** and **m** be **3**. Then we have

**-1 -2 -3** \+4 +5 +6 **-7 -8 -9** \+10 +11 +12

If **n = 4** and **m = 1**, then we have

**-1** \+2 **-3** \+4

Now your task is to find the summation of the numbers considering their signs.

Input

Input starts with an integer **T (≤ 10000)**, denoting the number of test cases.

Each case starts with a line containing two integers: **n** and **m (2 ≤ n ≤ 109, 1 ≤ m)**. And you can assume that **n** is divisible by **2\*m**.

Output

For each case, print the case number and the summation.

Sample Input

12 3

4 1

Sample Output

Case 1: 18

Case 2: 2

规律题，我开始还想着打表，后来发现规律了轻松ac

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #define LL long long
    using namespace std;
    //void init() {
    //	for(int i=1; i<n; i++) {
    //		a[i]=a[i-1]+i;
    //	}
    //}
    int main() {
    	int T,p=0;
    	scanf("%d",&T);
    	while(T--) {
    		LL n,m;
    		scanf("%lld%lld",&n,&m);
    		LL cnt=0;
    		cnt=n/(2*m)*m;
    		printf("Case %d: %lld\n",++p,cnt*m);
    	}
    	return 0;
    }

[Status]: http://acm.hust.edu.cn/vjudge/contest/129616#status//H/0
[Practice]: http://acm.hust.edu.cn/vjudge/problem/27004
[LightOJ 1294]: http://acm.hust.edu.cn/vjudge/problem/27004/origin
[12x12_uDebug.png]: http://acm.hust.edu.cn/vjudge/images/12x12_uDebug.png
[12x12_uDebug.png_uDebug]: https://www.udebug.com/LOJ/1294